cos^6 (pi/16) + cos^6 ((3pi)/16) + cos^6 ((5pi)/16) + cos^6 ((7pi)/16) = 5/4

1. Turning $\cos^6\theta$ into simpler pieces

The standard trick is to keep halving the power using

$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$

After two halvings and a bit of algebra we get

$\cos^6\theta = \frac{10 + 15\cos 2\theta + 6\cos 4\theta + \cos 6\theta}{32}$

This converts a power into a sum of cosines with multiple angles.

2. Why do we choose exactly those four angles?

The angles

$\theta \in \left\{\frac{\pi}{16},\; \frac{3\pi}{16},\; \frac{5\pi}{16},\; \frac{7\pi}{16}\right\}$

are equally spaced by $\frac{\pi}{8}$ inside the first quadrant. When we double, quadruple, or sextuple them, we get sets that wrap perfectly around the unit circle, making their cosines pair up with opposite signs and cancel out.

3. Adding everything

Write the required sum as

$S = \sum_{k=0}^{3} \cos^6\!\left(\frac{(2k+1)\pi}{16}\right)$

Plug the expansion from Step 1 into each term:

$S = \frac{1}{32}\Bigg[\underbrace{\sum 10}_{40} + 15\sum \cos 2\theta + 6\sum \cos 4\theta + \sum \cos 6\theta \Bigg]$

Because each grouped cosine sum equals zero (thanks to their symmetry), only the constant 40 survives, giving

$S = \frac{40}{32} = \frac{5}{4}$

That’s it!

What you learned

• Power-reducing identities change high powers of trig functions into multiple-angle cosines.
• Cleverly chosen angles can make most terms cancel, leaving only a simple constant.

What’s the problem?

We want to add four numbers that look scary:

• $\cos^6\!\left(\dfrac{\pi}{16}\right)$
• $\cos^6\!\left(\dfrac{3\pi}{16}\right)$
• $\cos^6\!\left(\dfrac{5\pi}{16}\right)$
• $\cos^6\!\left(\dfrac{7\pi}{16}\right)$

Big idea (in kid-speak)

Imagine turning a big Lego block (the power 6) into many small bricks (powers 1 and 2). When we break the block correctly, most bricks cancel each other when we add them up. Only a few simple bricks stay, giving a very tidy answer.

1. First, change $\cos^6\theta$ into a mix of $\cos(2\theta)$, $\cos(4\theta)$, $\cos(6\theta)$ and a plain number.
2. Add all four angles together. Because the angles are nicely spaced, the parts with $\cos(2\theta),\;\cos(4\theta),\;\cos(6\theta)$ cancel (they add up to zero).
3. What’s left is only the plain number part, which is super easy to add.

At the end, the scary sum is just $\dfrac{5}{4}$.

Step-by-step solution

1. Use the power-reducing identity

   $\cos^6\theta = \frac{10 + 15\cos 2\theta + 6\cos 4\theta + \cos 6\theta}{32}$

2. Write the desired sum

   $S = \cos^6\!\left(\frac{\pi}{16}\right) + \cos^6\!\left(\frac{3\pi}{16}\right) + \cos^6\!\left(\frac{5\pi}{16}\right) + \cos^6\!\left(\frac{7\pi}{16}\right)$

3. Insert the expansion into each term

   + \frac{1}{32}\Big[10 + 15\cos \tfrac{3\pi}{8} + 6\cos \tfrac{3\pi}{4} + \cos \tfrac{9\pi}{8}\Big] \\ + \frac{1}{32}\Big[10 + 15\cos \tfrac{5\pi}{8} + 6\cos \tfrac{5\pi}{4} + \cos \tfrac{15\pi}{8}\Big] \\ + \frac{1}{32}\Big[10 + 15\cos \tfrac{7\pi}{8} + 6\cos \tfrac{7\pi}{4} + \cos \tfrac{21\pi}{8}\Big]$$

4. Group like terms

   + 15\sum \cos 2\theta + 6\sum \cos 4\theta + \sum \cos 6\theta \Bigg]$$

5. Evaluate each cosine sum

   • $\sum \cos 2\theta = \cos \tfrac{\pi}{8}+\cos \tfrac{3\pi}{8}+\cos \tfrac{5\pi}{8}+\cos \tfrac{7\pi}{8}=0$
   • $\sum \cos 4\theta = \cos \tfrac{\pi}{4}+\cos \tfrac{3\pi}{4}+\cos \tfrac{5\pi}{4}+\cos \tfrac{7\pi}{4}=0$
   • $\sum \cos 6\theta = \cos \tfrac{3\pi}{8}+\cos \tfrac{9\pi}{8}+\cos \tfrac{15\pi}{8}+\cos \tfrac{21\pi}{8}=0$

6. Compute the final value

   $S = \frac{1}{32}\times 40 = \frac{40}{32} = \boxed{\dfrac{5}{4}}$